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3 years ago

An object speeding up, an object slowing down, and an object changing direction are all examples of acceleration.

Physics

1 answer:

emmainna [20.7K]3 years ago

6 0

<h2>Answer: True</h2>

When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down.

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A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and

mel-nik [20]

Answer:

7.35m/s²

Explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion

$\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\$

Fm is the moving force = 410N

$\mu$ is the coefficient of friction = 0.18

m is the mass = 45kg

g is the acceleration dur to gravity = 9.8m/s²

a is the acceleration of the wagon

Substitute the given data into the equation ang get ax

$Fm- \mu mg = ma_x\\410 - (0.18)(9.8)(45) = 45a_x\\410 - 79.38 = 45a_x\\330.62 = 45a_x\\a_x = 330.62/45\\a_x = 7.35m/s^2\\$

Hence the acceleration of the wagon is 7.35m/s²

8 0

3 years ago

Are moons 1-4 waxing are waning ?

solong [7]

Answer:

I want to help you but i cant.

Explanation:

please provide a screenshot or photos of moons 1-4

3 0

3 years ago

An atom with more or less neutrons than expected is a what ?

salantis [7]

Answer:

Isotope it will have a different number of neutrons than normal

4 0

2 years ago

HELLLP PLEASE || the graph below shows a conversion of energy for a skydive jumping out of a plane and landing safely on the gro

fenix001 [56]

Answer: I maybe wrong but i'm pretty sure its C) Kinetic energy

5 0

3 years ago

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

8 0

3 years ago